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    (Original post by Hunnybeebee)
    So:

    Halogenoalkane ------> primary amine

    Need excess ethanolic ammonia and heat

    Excess ammonia required for the displaced Cl- ion to react with ammonia to create NH4+Cl- from the HCl?

    Posted from TSR Mobile
    Yes up until the reagents and conditions.

    Excess nh3 is added so that there's more nh3 in the system than the primary amine made in the reaction. If there's more nh3 then it's more likely for the nh3 to react with the halogenoalkane than the primary amine. Of course you can't prevent primary amine from reacting with the halogenoalkane because it's a much better nucleophile than nh3 but to make sure you don't get as much secondary amines made you add excess nh3.
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    Do we have to know chiral synthesis etc in a lot of detail? If so could someone quickly give me an overview? Thanks!
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    (Original post by RetroSpectro)
    Can someone state all the points needed to describe the structure of Benzene and its bond angles?
    Benzene is made of 6 carbon atoms and 6 hydrogen atoms. Shape: planar hexagonal ring. Each carbon atom has 3 sigma bonds (bond angle of 120) and also one electron in a 2p orbital above and below the plane of the molecule. Sideways overlap of p orbitals creates the delocalised pi system.
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    (Original post by jerseyalevel)
    Do we have to know chiral synthesis etc in a lot of detail? If so could someone quickly give me an overview? Thanks!
    I don't think you need to know in detail. But it's just that chiral pool synthesis uses naturally occurring optical isomers (which is just one form of it so no racemic mixture etc.) in the synthetic route.
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    Hi guys?
    I am confused if mechanisms can be considered to be diagrams and hence can be drawn in pencils? I would appreciate if someone knows about it more or has asked the teacher and can shed more light on their point of view. Many thanks
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    (Original post by KB_97)
    Benzene is made of 6 carbon atoms and 6 hydrogen atoms. Shape: planar hexagonal ring. Each carbon atom has 3 sigma bonds (bond angle of 120) and also one electron in a 2p orbital above and below the plane of the molecule. Sideways overlap of p orbitals creates the delocalised pi system.
    you can called the 2p orbital above and below the plane 2pz. shape is trigonal planar for every carbon.
    6 resonance bonds for benzene if they ask for bonds.
    bond strength: single < resonance < double
    bond length: double < resonance < single
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    Do we need to know the monomers for Kevlar and terylene for example
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    just read the book, the page about synthetic methods of aliphatic compounds have a very misleading image, which lead to a lot of people thinking that carboxylic acids could not be reduced. amuses me that they do not include it in the spec, but give questions exactly about reducing carboxylic acids in the exam lol.
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    (Original post by lai812matthew)
    you can called the 2p orbital above and below the plane 2pz. shape is trigonal planar for every carbon.
    6 resonance bonds for benzene if they ask for bonds.
    bond strength: single < resonance < double
    bond length: double < resonance < single
    Really?

    I thought resonance (delocalised) bonds were stronger than double bonds?
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    (Original post by KB_97)
    Yes up until the reagents and conditions.

    Excess nh3 is added so that there's more nh3 in the system than the primary amine made in the reaction. If there's more nh3 then it's more likely for the nh3 to react with the halogenoalkane than the primary amine. Of course you can't prevent primary amine from reacting with the halogenoalkane because it's a much better nucleophile than nh3 but to make sure you don't get as much secondary amines made you add excess nh3.
    Do I have to know that in that much detail?

    So excess ammonia to react with any halogenoalkane present rather than the primary alkane to prevent forming secondary amines. Also reacts with HCl to form NH4+Cl-?

    Posted from TSR Mobile
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    (Original post by mil88)
    Really?

    I thought resonance (delocalised) bonds were stronger than double bonds?
    am just thinking reverse of bond length lol uncertain about this
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    what are the exact conditions of alkaline and acid hydrolysis
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    (Original post by lai812matthew)
    am just thinking reverse of bond length lol uncertain about this
    Well thinking about this logically, benzene is more stable than the kekule proposed model in that it requires certain condition to react with Br2, when the kekule one doesn't. Thus, more stable bonds, thus stronger.

    Well that's my thought process.
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    All we need to know about carbon-13 spectra (summed up):

    Although 12C is the most abundant isoptope of carbon, it has an even number of nucleons so cannot be used. Carbon-13 however has an odd number of nucleons so can be detected using low energy radio waves and it's possible to generate spectra. HOweveer, it's a LOT more difficult to detect than proton atoms because of it's low abundance and low magnetic moment (may ask us to compare the use of the two? lots more to mention if that is the case). 12C atoms DO interact with adjcaent protons but interactions are removed by decoupling so all absorptions appear as singlets. Each peak is a different carbon environment.
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    (Original post by Yazmin123)
    what are the exact conditions of alkaline and acid hydrolysis
    hot aqueous naoh or hot aqeous acid reflux
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    just thinking, they could ask us to suggest a suitable isotope for certain element (e.g. N,P) to perform a magnetic resonance.
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    (Original post by mil88)
    Well thinking about this logically, benzene is more stable than the kekule proposed model in that it requires certain condition to react with Br2, when the kekule one doesn't. Thus, more stable bonds, thus stronger.

    Well that's my thought process.
    thought benzene requires certain condition to react with Br2, is due to electrons being all delocalised and spread out around the ring, low pi electron density then kekule which have 3 double bonds, greater pi electron density blablabla......
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    (Original post by lai812matthew)
    just read the book, the page about synthetic methods of aliphatic compounds have a very misleading image, which lead to a lot of people thinking that carboxylic acids could not be reduced. amuses me that they do not include it in the spec, but give questions exactly about reducing carboxylic acids in the exam lol.
    but do they get reduced also by nabh4


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    Sorry i know this has been asked before:

    What would be the splitting pattern for CH3-CH2-CH2-CH3 (Butane)?

    I know there are 2 hydrogen environments due to the symmetry, but the symmentry also affects the splitting and I'm confused as to what the effect is exactly. Thanks in advance.
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    (Original post by ranz)
    but do they get reduced also by nabh4


    Posted from TSR Mobile
    no not by nabh4, but another molecule we don't need to know
 
 
 
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